Synthesis of Propane via Major Product Organic Reactions

To determine which of the given reactions yield propane ($\text{CH}_3\text{CH}_2\text{CH}_3$) as a major product, we analyze each reaction option step-by-step:

Option (A): Soda-Lime Decarboxylation

• Reactant: Sodium salt of butanoic acid, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}$ (Sodium butanoate).
• Reagent: Soda-lime ($\text{NaOH} + \text{CaO}$) and heat ($\Delta$).
• Reaction: $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{CH}_3\text{CH}_2\text{CH}_3 + \text{Na}_2\text{CO}_3$
• Explanation: Heating the sodium salt of a carboxylic acid with soda-lime causes decarboxylation, removing the carboxylate group as sodium carbonate ($\text{Na}_2\text{CO}_3$). This produces an alkane containing one less carbon atom than the parent carboxylate salt. Starting with 4-carbon sodium butanoate yields 3-carbon propane as the major product.
• Result: Produces propane.

---

Option (B): Kolbe's Electrolysis

• Reactant: Aqueous solution of sodium butanoate, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + \text{H}_2\text{O}$.
• Reaction: $2\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^- \xrightarrow{\text{Electrolysis}} \text{CH}_3\text{CH}_2\text{CH}_2-\text{CH}_2\text{CH}_2\text{CH}_3 + 2\text{CO}_2 + 2e^-$
• Explanation: Electrolysis of an aqueous solution of sodium carboxylate leads to decarboxylation at the anode to form alkyl free radicals ($\text{CH}_3\text{CH}_2\text{CH}_2^\bullet$). Two such propyl radicals dimerize to yield a symmetrical alkane, $n$-hexane ($\text{C}_6\text{H}_{14}$), as the major product.
• Result: Does NOT produce propane.

---

Option (C): Reduction of Monohalide with Zinc and Acid

• Reactant: 1-chloropropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$).
• Reagent: Zinc and dilute hydrochloric acid ($\text{Zn}, \text{dil. HCl}$).
• Reaction: $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{Zn} + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{CH}_3 + \text{Zn}^{2+} + \text{Cl}^-$
• Explanation: Treatment of alkyl halides with zinc metal in an acidic medium leads to reduction of the carbon-halogen bond to form the corresponding alkane. Here, 1-chloropropane is reduced directly to propane.
• Result: Produces propane.

---

Option (D): Reduction of Geminal Dihalide

• Reactant: 2,2-dibromopropane ($\text{CH}_3\text{CBr}_2\text{CH}_3$).
• Reagent: Zinc ($\text{Zn}$).
• Reaction: $\text{CH}_3\text{CBr}_2\text{CH}_3 \xrightarrow{\text{Zn, reduction}} \text{CH}_3\text{CH}_2\text{CH}_3$
• Explanation: Geminal dihalides (gem-dihalides) undergo complete reduction upon treatment with active metals like zinc to form the corresponding alkane, replacing both halogen atoms with hydrogen atoms to produce propane.
• Result: Produces propane.

---

Conclusion:

The reactions in options A, C, and D produce propane as the major product.

Correct Options: A, C, D